If A + B + C = pi , then cos , 2A + cos , 2B | vedclass

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(C) Given $A + B + C = \pi$,so $A + B = \pi - C$.Using the sum-to-product formula: $\cos 2A + \cos 2B = 2 \cos(A + B) \cos(A - B)$.Also,$\cos 2C = 2 \

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$- 1 - 4 \cos A \cos B \cos C$

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(C) Given $A + B + C = \pi$,so $A + B = \pi - C$.Using the sum-to-product formula: $\cos 2A + \cos 2B = 2 \cos(A + B) \cos(A - B)$.Also,$\cos 2C = 2 \cos^2 C - 1$.Substituting these into the expression:$L.H.S. = 2 \cos(A + B) \cos(A - B) + 2 \cos^2 C - 1$.Since $\cos(A + B) = \cos(\pi - C) = - \cos C$,we have:$L.H.S. = 2(-\cos C) \cos(A - B) + 2 \cos^2 C - 1$.$L.H.S. = - 1 - 2 \cos C [\cos(A - B) - \cos C]$.Since $\cos C = - \cos(A + B)$,we substitute:$L.H.S. = - 1 - 2 \cos C [\cos(A - B) + \cos(A + B)]$.Using $\cos(A - B) + \cos(A + B) = 2 \cos A \cos B$:$L.H.S. = - 1 - 2 \cos C [2 \cos A \cos B] = - 1 - 4 \cos A \cos B \cos C$.

If $A + B + C = \pi ,$ then $\cos \, 2A + \cos \, 2B + \cos \, 2C = $

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